Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Appendix A - A.4 - Quadratic Equations - Exercises - Page 550: 40

Answer

The three real solutions are: $x = 0,~~$ $~~x=1,~~$ $~~x=6$

Work Step by Step

Let's consider a trinomial in this form: $~x^2+bx+c$ To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = c$ Then the next step is to rewrite the trinomial as follows: $~x^2+bx+c = (x+r)~(x+s)$ We can use the GCF to factor the given equation: $4x^3-28x^2+24x = 0$ $(4x)~(x^2-7x+6) = 0$ To factor the left side of this equation, we need to find two numbers $r$ and $s$ such that $r+s = -7~$ and $~r\times s = 6$. We can see that $(-1)+(-6) = -7~$ and $(-1)\times (-6) = 6$ We can solve the equation as follows: $4x^3-28x^2+24x = 0$ $(4x)~(x^2-7x+6) = 0$ $(4x)~(x-1)~(x-6) = 0$ $4x = 0,~~$ $~~x-1=0,~~$ or $~~x-6=0$ $x = 0,~~$ $~~x=1,~~$ or $~~x=6$
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